NEET physics
Thermal Properties of Matter
Specific Heat Capacity
- Specific heat is defined as the amount of heat per unit mass absorbed or rejected by the substance to change its temperature by one.
Mathematicallycan be written as:-
Where
- ΔQ = amount of heat absorbed or rejected by a substance
- m = mass
- ΔT = temperature change
- It depends on the nature of the substance and its temperature.
- The SI unit of specific heat capacity is J kg–1 K–1.
Molar specific heat capacity: -
- Heat capacity per mole of the substance is the defined as the amount of heat (in moles) absorbed or rejected(instead of mass m in kg) by the substance to change its temperature by one unit.
Mathematically can be written as:-
C = S/ μ= ΔQ / μ ΔT
Where
- μ= amount of substance in moles
- C = molar specific heat capacity of the substance.
- ΔQ = amount of heat absorbed or rejected by a substance.
- ΔT = temperature change
It depends on the nature of the substance and its temperature. The SI unit of molar specific heat capacity is Jmol–1 K–1
Molar specific heat capacity (Cp):-
- If the gas is held under constant pressure during the heat transfer, then the corresponding molar specific heat capacity is called molar specific heat capacity at constant pressure (Cp).
Molar specific heat capacity (Cv):-
- If the volume of the gas is maintained during the heat transfer, then the corresponding molar specific heat capacity is called molar specific heat capacity at constant volume (Cv).
- Water has highest specific heat of capacity because of which it is used as a coolant in automobile radiators and in hot water bags.
Problem:- In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150 °C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27 °C. The final temperature is 40 °C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your Solution greater or smaller than the actual value for specific heat of the metal?
Solution:- Mass of the metal, m = 0.20 kg = 200 g
Initial temperature of the metal, T1 = 150°C
Final temperature of the metal, T2 = 40°C
Calorimeter has water equivalent of mass, m’ = 0.025 kg = 25 g
Volume of water, V = 150 cm3
Mass (M) of water at temperature T = 27°C: 150 × 1 = 150 g
Fall in the temperature of the metal:
ΔT = T1 – T2 = 150 – 40 = 110°C
Specific heat of water, Cw = 4.186 J/g/°K
Specific heat of the metal = C
Heat lost by the metal, θ = mCΔT … (i)
Rise in the temperature of the water and calorimeter system:
ΔT = 40 – 27 = 13°C
Heat gained by the water and calorimeter system:
Δθ’’ = m1 CwΔT’
= (M + m′) Cw ΔT’ … (ii)
Heat lost by the metal = Heat gained by the water and calorimeter system
mCΔT = (M + m’) Cw ΔT’
200 × C × 110 = (150 + 25) × 4.186 × 13
If some heat is lost to the surroundings, then the value of C will be smaller than theactual value.
NEET Thermal Properties of Matter NCERT Chapter 11 Free Notes for Best Revision
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